Rossi energy catalyst – a big hoax or new physics?

[DancingFool]

Might his "catalyst" be the same thing as his "heater" which is a SPP generator? If so, then the initiating particle would be a neutron, not a proton, and INITIATING a nuclear reaction would be simple.

Again, nope.

His "catalyst" is claimed to be a simple compound, no exotic or radioactive elements. His heater is claimed to be a simple resistive heater. And no neutrons.

[DancingFool]

Interesting. Perhaps it is just a language difference, but I am under the impression that any time one brings H into contact with the surface of Ni, especially under heat, the H enters into surface micro-cracks and finally works its way into the lattice itself. Given nano particles, the amount of surface micro-cracking would be huge, accelerating the diffusion into the lattice. This would load the lattice without a long duration "loading process". ICBW.

Sorry for the confusion. I was referring to the loading claimed for P/F processes, where deuterons have to permeate the palladium matrix. Nickel will, of course, behave as you say, but since the initiation temperature is claimed to be as low as 150 C, it seems unlikely.

[KitemanSA]

What is an atypical gamma?

A gamma that is not what what you would typically get. But I didn't say atypical, you did. In this case, a "typical" gamma was ~6MeV by CMB's statement (IIRC). If 99% of the energy was shed a different way and only 0.06MeV were left...

What is an SPP?

As I wrote earlier, surface plasmon polariton. Did I remember the term wrong? Entirely possible.

HPA.

No idea. Didn't use it.

[KitemanSA]

Might his "catalyst" be the same thing as his "heater" which is a SPP generator? If so, then the initiating particle would be a neutron, not a proton, and INITIATING a nuclear reaction would be simple.

Again, nope.

His "catalyst" is claimed to be a simple compound, no exotic or radioactive elements. His heater is claimed to be a simple resistive heater. And no neutrons.

I have read so many DIFFERENT statements about his "catalyst" that I am not sure which, if any, to believe. Indeed, I was under the impression that he had recently said it was NOT a chemical additive.

Note he HAS a simple resistance heater too. It is located at the bottom of the unit. It would not surprise me if he were making obfuscatory statements to delay or misdirect any efforts to replicate.

[MSimon]

Kiteman,

The "not typical" (atypical) gammas you refer to (.06 MeV) are called X-Rays. In fact low energy X-rays.

[chrismb]

Generally, the definition of gammas are EM from a nucleus, whereas X-rays are EM from electron excitations [viz. man made]. So gammas can actually be lower energy than X-rays. The difference is their origin.

What would be anomalous would be multiple gamma emissions from a single excited atom. Now that would be interesting, but there's no indication of this in the Rossi-tossi.

[Giorgio]

I am wondering whether WL's surface plasmon polariton(sp?) thingees would spread across all the surface area of the Ni particles if they were created on the inner surfaces of the cylinder of Ni powder. I believe I read that Rossi worked with nano particles which are the bases of quantum dot lasers, no? Might his "catalyst" be the same thing as his "heater" which is a SPP generator? If so, then the initiating particle would be a neutron, not a proton, and INITIATING a nuclear reaction would be simple.

I doubt you can get SPP out of simply heating. My knowledge was that the type of energy intensities needed to get SPP was obtainable only with laser illumination of the surface. I might be wrong thought.

If for the sake of discussion we have to take for goods his word that there is a catalyst than probably we should try to consider it as a real catalyst (i.e. a chemical compound that acts as a mediator in the process creating one or more intermediate steps) and not simply something that allow a direct fusion of H with Ni.
When hearing various interview to Focardi his choice of words was more consistent to what I mentioned above than to a simple "primer" to ignite the fusion reaction.
As I said, just personal impressions, so take them for what they are worth.

[Giorgio]

What is an SPP?

http://www.dep.anl.gov/postdocs/symposi ... ERIALS.pdf

[MSimon]

chris,

Evidently not only am I Old Navy but I'm also Old Physics definitions:

The distinction between X-rays and gamma rays has changed in recent decades. Originally, the electromagnetic radiation emitted by X-ray tubes had a longer wavelength than the radiation emitted by radioactive nuclei (gamma rays).[5] Older literature distinguished between X- and gamma radiation on the basis of wavelength, with radiation shorter than some arbitrary wavelength, such as 10−11 m, defined as gamma rays.[6] However, as shorter wavelength continuous spectrum "X-ray" sources such as linear accelerators and longer wavelength "gamma ray" emitters were discovered, the wavelength bands largely overlapped. The two types of radiation are now usually distinguished by their origin: X-rays are emitted by electrons outside the nucleus, while gamma rays are emitted by the nucleus.[5][7][8][9]

http://en.wikipedia.org/wiki/X-ray

[KitemanSA]

Kiteman,

The "not typical" (atypical) gammas you refer to (.06 MeV) are called X-Rays. In fact low energy X-rays.

I.e., not your typical gamma, right?

[KitemanSA]

I am wondering whether WL's surface plasmon polariton(sp?) thingees would spread across all the surface area of the Ni particles if they were created on the inner surfaces of the cylinder of Ni powder. I believe I read that Rossi worked with nano particles which are the bases of quantum dot lasers, no? Might his "catalyst" be the same thing as his "heater" which is a SPP generator? If so, then the initiating particle would be a neutron, not a proton, and INITIATING a nuclear reaction would be simple.

I doubt you can get SPP out of simply heating. My knowledge was that the type of energy intensities needed to get SPP was obtainable only with laser illumination of the surface. I might be wrong thought.

Please look at the highlighted portion of my quote. The thought is that MAYBE his "internal heater" does more than just heat. If there are QD lasers built into it...

Please note that I am having fun speculating. I am not stating that I BELIEVE that what I have suggested is the case, merely that, AFAIK, it is not outside what I understand physics to allow. If it turned out that my speculations were true, well, I'd be flabbergasted that I was right, but it would in no way upset my understanding of physics and the universe it claims to portray.

[Giorgio]

I am wondering whether WL's surface plasmon polariton(sp?) thingees would spread across all the surface area of the Ni particles if they were created on the inner surfaces of the cylinder of Ni powder. I believe I read that Rossi worked with nano particles which are the bases of quantum dot lasers, no? Might his "catalyst" be the same thing as his "heater" which is a SPP generator? If so, then the initiating particle would be a neutron, not a proton, and INITIATING a nuclear reaction would be simple.

I doubt you can get SPP out of simply heating. My knowledge was that the type of energy intensities needed to get SPP was obtainable only with laser illumination of the surface. I might be wrong thought.

Please look at the highlighted portion of my quote. The thought is that MAYBE his "internal heater" does more than just heat. If there are QD lasers built into it...

Please note that I am having fun speculating. I am not stating that I BELIEVE that what I have suggested is the case, merely that, AFAIK, it is not outside what I understand physics to allow.

An internal laser? While possible I highly doubt.
Lasers normally have a pretty tight range of temperature of operation, and the internal temperatures of the reactor should prevent it from working correctly.
Than, again, without any data all is possible and is hard to say anything.

[KitemanSA]

An internal laser? While possible I highly doubt.
Lasers normally have a pretty tight range of temperature of operation, and the internal temperatures of the reactor should prevent it from working correctly.
Then, again, without any data all is possible and is hard to say anything.

Yes, I have read many statements of the operating temperature of this thing. 60, 600, 1600! Typos? INTENTIONAL typos? Who knows.

How hot does a QD laser still function at? Can the laser function thru a water bath in a protected glass cylinder?

Well, more and more afield in speculations.

Wish we could get some Polywell (no wait, we are talking Rossi here) data.

[Giorgio]

How hot does a QD laser still function at? Can the laser function thru a water bath in a protected glass cylinder?

I really have no idea. I presume the main issues of an increase in temperature will be wavelength modification and divergence parameters.
They normally tend to get out of optimal values pretty quickly.
Maybe someone else can give a decent reply on this.

[D Tibbets]

Note : If the protons (hydrogen nuclei) are not being fused into heavier elements like deuterium, helium, etc. that are lighter than iron, but it is being added to nickel- directly or through a conversion to a neutron intermediate, these are generally endothermic reactions.

Code: Select all

Weights in Isotopic Mass (u)
63Cu  = 62.9295975
62Ni  = 61.9283451
        -------------
Delta =  1.0012524 

Proton=  1.00782503207

The proton weighs MORE than the delta between 62Ni and 63Cu. The difference is converted into energy. The reaction is EXOthermic. Please remember that. This is the second time folks on this forum have got it wrong. It is just that 62Ni is more stable than 63Cu so the conversion is not likely to stick. TYPICALLY, if a Ni absorbes a proton, it will just spit it back out again. Seems something has to interrupt that "spit it out" process to make the proton stick. (Same with a neutron I believe).

This solution bothered me. Was N1 62 to Cu63 an exception to the binding energy limits? It appears so when looking at the mass/ energy difference comparison. But after some research, I conclude that this is a misunderstanding on your part. As pointed out in the quote below, any nucleus will have a shortfall between it's mass and the mass of its constituents. This difference is the binding energy. This ratio maximizes at iron- thus it's position at the peak of the binding energy curve.


http://hyperphysics.phy-astr.gsu.edu/hb ... ucbin.html

"Nuclei are made up of protons and neutron, but the mass of a nucleus is always less than the sum of the individual masses of the protons and neutrons which constitute it. The difference is a measure of the nuclear binding energy which holds the nucleus together."

Thus when building elements heavier than Fe this binding energy / mass ratio drops. Another way too look at is that while the total mass/ energy in heavier elements is greater, the harvestable energy through nuclear processes is decreassed. Your decreassed mass of Cu63 comparred to Ni62 + proton does not mean that you have released free energy (exthermic), it means that the cupper63 has less total energy (E=MC^2) than the Ni62 +proton. ie: you had to put energy into the system to make the atom.
If you annialate a Cu53 nucleus with an anti Cu63 nucleus you would get more energy that from annialating a Ni62 nucleus. But when you add the proton- anti proton annialation to the Ni62 , the total energy released would be greater (due to the mass difference represented by the decreased binding energy of the Cu63.

It is complex and I can easily become confused, but the simple answer is that no , the Ni62 +P to Cu63 is not exothermic!, it is endothermic (you have to put enenergy into the system to drive the reaction)*. This is a hard fact of physics. To overcome this in the Rossi claimed reaction, you not only have to crerate new physics, but ignor or change this basic understanding of nuclear binding energy.



Some other links describing neucleosynthesis:


http://kencroswell.com/Copper.html

http://iopscience.iop.org/0004-637X/710/2/1557

http://en.wikipedia.org/wiki/Stellar_nucleosynthesis

http://en.wikipedia.org/wiki/Nucleosynthesis

* The annihilation example is not clear in my mind, it may be misleading. Another way of describing it is that you have a bag of protons and neurtons- this energy- mass is unchanging and depends only on their numbers. This energy is irrelevant in terms of the energy needs to build or break down these nucleii. The pertinent energy is the binding energy. Hydrogen has the least binding energy, iron has the most, Metals heavier than iron again have lower binding energies. This translates to any reaction going towards iron can release binding energy, those going away can only consume energy.

ie: the available binding energy in a Cu63 nucleus may be more than the binding energy of a Ni62 nucleus, but it is less than the binding energy of a Ni62 nucleus and a proton added together- less total available energy, so energy has to be added to the system to create the Cu63. This energy is represented by the difference in binding energy, or in your example the mass differences. When you use the mass of the proton it is it's true rest mass, but when you have the proton in a nucleus with other protons and neutrons, it's apparent mass will be less. The difference is the binding energy. I don't know how neutrons fit into this picture. I suspect that they are irrelevant for this narrow discussion. The binding energy is involved with the strong force, while neutrons are only involved with the weak force(?).
It is difficult to conceive of a proton having a binding energy when it is not actually bound to anything. This is essentailly a binding energy of ...f 1 ( I started to say zero, but I think that would represent pure energy). There is a lot of territory between the binding energy of a proton, perhaps represented by the glueons that binds the quarks together, and pure energy. Perhaps using other reactions where there is more than one nuclear component is more revealing (like deuterium or above reactions). A useful viewpoint is to turn the typical nuclear binding graphs (like the one in the first link) upside down. Then the graph would be showing the amount of energy releasable as opposed to the amount of energy retained. And remember you are not comparing the releasable energy between Cu63 and Ni62, but the total of Ni62 plus hydrogen. The heavier elements will release energy compared to iron if they fission or lose mass through some other radioactive process, but you need to remember that it took as much or more energy to create these heavier elements in the first place. Also remember that isotopes can contain excess energy, at least temporarily in the form of excited isomers. But again, these isomers had to be created in the first place through energy input.


Dan Tibbets